# Find the maximum volume of an open box made from 3ft by 8ft rectangular piece of sheet metal by cutting out squares of equal sizes of four corners Find the maximum volume of an open box made from a 3' by 8' rectangular sheet with equal squares cut fromt he 4 corners:

The volume is `V=(3-2x)(8-2x)x`

** Each length has a segment of length x cut from both corners; the height of the box will be x **

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Find the maximum volume of an open box made from a 3' by 8' rectangular sheet with equal squares cut fromt he 4 corners:

The volume is `V=(3-2x)(8-2x)x`

** Each length has a segment of length x cut from both corners; the height of the box will be x **

To find the maximum we take the derivative with respect to x.

`V=4x^3-22x^2+24x` after expanding so:

`(dV)/(dx)=12x^2-44x+24=4(3x^2-11x+6)=4(3x-2)(x-3)`

A maximum of a continuous function can only occur at an critical point (where `f'(x)=0` or `f'(x)` fails to exist. Since the derivative is a polynomial it is everywhere continuous).

Setting `(dV)/(dx)=0` we get `4(3x-2)(x-3)=0==>x=2/3` or `x=3`

`x=3` does not make sense in the context of the problem. (After cutting squares of length 3 from the side of length 3, we have a negative side length) so x=`2/3` .

If x=`2/3` the volume of the box is `V=(3-4/3)(8-4/3)(2/3)=200/27`

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The maximum volume is `200/27~~7.41` cu ft

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